If the sum of the series 20 + 19[3 5] + 19[1 5] + 18[4 5] + ...

Question

If the sum of the series

20 + 19$${3 \over 5}$$ + 19$${1 \over 5}$$ + 18$${4 \over 5}$$ + ...

upto nth term is 488
and the nth term is negative, then :

Accepted Answer

$$S = {{100} \over 5} + {{98} \over 5} + {{96} \over 5} + {{94} \over 5} + ...\,n$$

$$\,{S_n} = {n \over 2}\left( {2 \times {{100} \over 5} + (n - 1)\left( {{{ - 2} \over 5}} \right)} \right) = 488$$

$$ \Rightarrow $$ $$n(100 - n + 1) = 488 \times 5$$

$$ \Rightarrow $$ $${n^2} - 101n + 488 \times 5 = 0$$

$$ \Rightarrow $$ $$n = 61,\,40$$

For negative term n = 61

$$\,{T_n} = a + (n - 1)d = {{100} \over 5} - {2 \over 5} \times 60$$

$= 20 - 24 =  - 4$

Solutions · 1